Question

help me with the steps.
Find the center, transverse axis, vertices, foci, and asymptotes.
x^2-9y^2-4x-72y-144=0

Answer 1

You must complete the square in x and y, separately.

(x^2 - 4x + ____) - 9(y^2 + 8y + ____) = 144 + ____ + 9(____)

Answer 2

so they will look like

(x^2 - 4x + 4) -9(y^2 + 8y + 16 ) =144 + 4 + 144

Answer 3

@tkhunny

Answer 4

That was very good. Did you need me to do all that setup or could you have done that yourself?

Answer 5

didnt know how to put it in the fraction form and also forgot that need to add to the sum too. so how do I put in the form to find the center?

Answer 6

btw That should have been NEGATIVE 9 on the right hand side. Sorry about that.

144 + 4 - 144 = 4

Answer 7

Rewrite the two perfect square trinomials into their square binomial forms and divide by 4. You'll almost be there.

Answer 8

so it will look like (x-2)^2 / 4 - 9(y+4)^2 / 4 =1

Answer 9

and the center would be (2, -4)

Answer 10

but not sure if it would be transverse on the x or y

Answer 11

@tkhunny

Answer 12

Write it like this.

\(\dfrac{(x-2)^{2}}{4} - \dfrac{(y+4)^{2}}{4/9} = 1\)

Answer 13

so it will make it transverse on the x

Answer 14

1) Wipe out the entire y-term. Is there a solution for x? Yes. This gives the direction and location of the vertices.

Center: (2,-4)
Vertex: (2 - 2,-4) and (2 + 2,-4)

2) On the other hand, wipe out the entire x-term. Is there a solution for y? No. This gives the direction and location of the "other" vertices.

Center: (2,-4)
Vertex: (2,-4 + 2/3) and (2,-4 - 2/3)

Answer 15

so its transverse on the x since there is a solution for x and not for y

Answer 16

Yes.

Answer 17

so the vertex you had how did you get them.

Answer 18

In that final form, the length of the transverse axis is pretty obvious. 4 = 2^2. It's that base 2 that's important. That's half the length of the transverse axis. That's why I inserted "+2" and "-2" to identify the vertices.

Answer 19

oh okay that just confused me. to find the foci I would just need to do c^2 = a^2 +b^2 right so it will be c^2 = 4 + 4/9 and it will make c^2 = 24/5 and c will equal to sqroot of 24/5

Answer 20

?? Better try that addition again. 4 + 4/9

Answer 21

oops i mean 40/9 for c^2

Answer 22

:-)

Answer 23

so the foic will be (-sqroot 40/9, -4) and sqroot 40/9, -4)

Answer 24

1) I would prefer (2/3)sqrt(10)
2) No, that is (h+c,k) (h-c,k) You're 'h' is missing.

Answer 25

so it will be (2-(2/3)sqroot 10, -4) and 2+(2/3)sqroot 10, -4)

Answer 26

I'd go with that! Good work.

Asymptotes?

Answer 27

to find asymtptes is it y=a/b x and y= -a/b x so it will be +- 3

Answer 28

Not quite. h and k are missing again. Those are good only for a center at the Origin.

(y-k) = (b/a)(x-h)

Ringing bells? One would also need to remember which way 'a' goes.

You can just do it yourself, without the silly formula. One point on the line is (h,k) and the slope is, indeed, b/a. Just write it down from the point-slope form of a line.

Answer 29

so its nto a/b but b/a and what is x?

Answer 30

not*

Answer 31

If 'a' is horizontal, it's b/a.
If 'a' is vertical, it's a/b.
'x' is part of the equation. Are you trying to replace it?

Answer 32

oh no! nvm just notice that haha. and okay i understand now! thank you so MUCH for taking your time and helping me!